Let X be symmetric matrix and X=USV^T its SVD decomposition. If X is positive semi-definite, SVD is the same as eigenvalue decomposition (with appropriate choice of matrix U, and order of eigenvalues), so the statement U=V holds. However, I think that the statement is not true if X is not positive semi-definite. For example, for SVD decomposition of matrix X = diag(-2,-1) U = -V. I think that for matrices X that are not positive semi-definite, abs(U)=abs(V), but signs differ for eigenvectors that correspond to negative eigenvalues. In the solution it is used an argument that U and V consist of eigenvectors of the same matrix, but that still does not tell us anything about the sign.
Also, why are we showing that S has all non-negative entries if X is positive semi definite, since it is always true?
Thanks for the question, and sorry for the confusion. I remember discussing the question with TAs just before the session. We came to the same conclusion as you.
The question seems to hold a slightly looser definition of SVD. I think the statement in the question should have been formulated as:
For a symmetric X, you can write X = U S U^T, for U orthogonal and S diagonal.
For an s.p.d. X, you can write X = U S U^T, for U orthogonal and S diagonal with non-negative entries.
Because 'the SVD' is not unique, especially not if the definition is used very loosely (allowing negative entries in S), it's not so accurate to call those decompositions 'the SVD'.
Lab 12, theory questions - problem 2
Let X be symmetric matrix and X=USV^T its SVD decomposition. If X is positive semi-definite, SVD is the same as eigenvalue decomposition (with appropriate choice of matrix U, and order of eigenvalues), so the statement U=V holds. However, I think that the statement is not true if X is not positive semi-definite. For example, for SVD decomposition of matrix X = diag(-2,-1) U = -V. I think that for matrices X that are not positive semi-definite, abs(U)=abs(V), but signs differ for eigenvectors that correspond to negative eigenvalues. In the solution it is used an argument that U and V consist of eigenvectors of the same matrix, but that still does not tell us anything about the sign.
Also, why are we showing that S has all non-negative entries if X is positive semi definite, since it is always true?
Is there anything that I misunderstood? Thanks!
Hi Jana,
Thanks for the question, and sorry for the confusion. I remember discussing the question with TAs just before the session. We came to the same conclusion as you.
The question seems to hold a slightly looser definition of SVD. I think the statement in the question should have been formulated as:
Because 'the SVD' is not unique, especially not if the definition is used very loosely (allowing negative entries in S), it's not so accurate to call those decompositions 'the SVD'.
Thank you!
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